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Answer:
pH = 11.95≈12
Explanation:
Remember the reaction among aqueous acetic acid (
) and aqueous sodium hydroxide (NaOH)

First step. Need to know how much moles of the substances are present
= 0.0025 mol NaOH
0.003 mol NaOH *
/ 1 mol NaOH = 0.003 mol CH_3COOH[/tex]
NaOH is in excess. Now, how much?
0.003 mole NaOH - 0.0025 mole NaOH = 0.0005 mole NaOH
Then, that amount in excess would be responsable for the pH.
Third step. Know the pH
Remember that pH= -log[H+]
According to the dissociation of water equilibrium
Kw=[H+]*[OH-]= 10^(-14)
The dissociation of NaOH is
NaOH -> 
Now, concentration of OH^{-}[/tex] would be given for the excess of NaOH.
[OH-]= 0.0005 mole / 0.055 L = 0.00909 M
Careful: we have to use the total volumen
Les us to calculate pH
![pH= -log [H+]\\pH= -log \frac{K_w}{[OH-]} \\pH= 11.95](https://tex.z-dn.net/?f=pH%3D%20-log%20%5BH%2B%5D%5C%5CpH%3D%20-log%20%5Cfrac%7BK_w%7D%7B%5BOH-%5D%7D%20%5C%5CpH%3D%2011.95)
Answer:
B. Composed of molecules relatively far apart.
Explanation:
The gas we call "air" has molecules that are relatively far apart.
Answer:
The
solution has a higher osmotic pressure and higher boiling point than LiCl solution.
Explanation:
As concentrations of two aqueous solutions are same therefore we can write:
,
and 
where
,
and
are lowering of vapor pressure, elevation in boiling point and osmotic pressure of solution respectively.
is van't hoff factor.
= total number of ions generated from dissolution of one molecule of a substance (for strong electrolyte).
Here both
and LiCl are strong electrolytes.
So,
and 
Hence, lowering of vapor pressure, elevation in boiling point and osmotic pressure will be higher for
solution.
Therefore the
solution has a higher osmotic pressure and higher boiling point than LiCl solution.